Certainly! Let's break down the problem step-by-step:
### 6.1.1 Define the term one mole of a substance.
One mole of a substance is the amount of that substance that contains [tex]\(6.022 \times 10^{23}\)[/tex] entities (such as atoms, molecules, or ions). This number is known as Avogadro's number. It is defined so that one mole of carbon-12 atoms has a mass of exactly 12 grams.
### 6.1.2 Balance the equation for the reaction.
The unbalanced chemical equation for the reaction between sodium carbonate ([tex]\(Na_2CO_3\)[/tex]) and hydrochloric acid ([tex]\(HCl\)[/tex]) is:
[tex]\[
Na_2CO_3(s) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l) + CO_2(g)
\][/tex]
To balance this equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. The balanced equation is:
[tex]\[
Na_2CO_3(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)
\][/tex]
### 6.1.3 Calculate the mass of sodium carbonate that reacted.
We need to perform the following steps:
1. Convert the volume of [tex]\(CO_2\)[/tex] from cm[tex]\(^3\)[/tex] to dm[tex]\(^3\)[/tex]:
[tex]\[
\text{Volume of } CO_2 = 306 \, \text{cm}^3 = 0.306 \, \text{dm}^3
\][/tex]
2. Calculate the number of moles of [tex]\(CO_2\)[/tex] produced:
The molar volume of a gas at room temperature is given as [tex]\(24.45 \, \text{dm}^3/\text{mol}\)[/tex].
[tex]\[
\text{Number of moles of } CO_2 = \frac{\text{Volume of } CO_2}{\text{Molar volume}} = \frac{0.306 \, \text{dm}^3}{24.45 \, \text{dm}^3/\text{mol}} = 0.012515337423312884 \, \text{mol}
\][/tex]
3. Link moles of [tex]\(CO_2\)[/tex] to moles of [tex]\(Na_2CO_3\)[/tex]:
According to the balanced chemical equation, 1 mole of [tex]\(Na_2CO_3\)[/tex] produces 1 mole of [tex]\(CO_2\)[/tex]. Therefore, the moles of [tex]\(Na_2CO_3\)[/tex] that reacted are equal to the moles of [tex]\(CO_2\)[/tex] produced:
[tex]\[
\text{Moles of } Na_2CO_3 = 0.012515337423312884 \, \text{mol}
\][/tex]
4. Calculate the mass of [tex]\(Na_2CO_3\)[/tex] that reacted:
The molar mass of [tex]\(Na_2CO_3\)[/tex] (sodium carbonate) is [tex]\(105.99 \, \text{g/mol}\)[/tex].
[tex]\[
\text{Mass of } Na_2CO_3 \text{ reacted} = \text{Moles of } Na_2CO_3 \times \text{Molar mass of } Na_2CO_3 = 0.012515337423312884 \, \text{mol} \times 105.99 \, \text{g/mol} = 1.3265006134969326 \, \text{g}
\][/tex]
### Final Results:
- Volume of [tex]\(CO_2\)[/tex]: [tex]\(0.306 \, \text{dm}^3\)[/tex]
- Moles of [tex]\(CO_2\)[/tex]: [tex]\(0.012515337423312884 \, \text{mol}\)[/tex]
- Mass of [tex]\(Na_2CO_3\)[/tex] reacted: [tex]\(1.3265006134969326 \, \text{g}\)[/tex]
This completes the detailed, step-by-step solution for the problem.
